$T^\mu{}_\mu = 0$? If this is the case then I get the feeling that if the indices are doing things like that, i.e., if you have repeated indices on the tensor, then you will end up with a scalar - and maybe that that scalar will be zero.

neither of which necessarily vanish. (They are physically important because they enter into Einstein's field equation of general relativity: $ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 8\pi G T_{\mu\nu}$).

I think that $g_{\nu\sigma}g^{\mu\sigma} = \delta^\nu_\mu$ and $\frac{p^\mu n_\nu + p_\nu n^\mu}{p \cdot n}$ I think, is simplified as far as it can go.

I think you mean $\delta^\mu_\nu$, which is numerically equal to $\delta^\nu_\mu$, but still different. :) Otherwise you're correct.

Here's where my understanding breaks down, how do I evaluate

$g_{\mu\nu} p^\mu n^\nu$ in terms of raising and lowering, which index do I choose? It must be a scalar, since it is a double sum. $g_{\mu\nu} p^\mu n^\nu = p \cdot n$.

Either index - you get the same answer both ways. Your answer is right: $p\cdot n$.

does the position of the index matter? Is there a distinction between $g_\mu^\nu$, $g^\nu{}_\mu$ and $g^\mu{}_\nu$ for any tensor? I can see where I have overlooked the multiplication of the very last term you mentioned now, that makes sense. But with $g^\mu{}_\mu$ I just thought we would have $-1+1+1+1$, but actually we have $g_{\mu\nu}g^{\mu\nu}$ is this $ \left( \begin{array}{c} -1 \\ 1 \\ 1 \\ 1 \end{array} \right)\left( \begin{array}{cccc} -1 & 1 & 1 & 1 \end{array} \right) $?
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shilovMar 14 '13 at 10:40

1

In general there is a difference between say $A_{\ \ \mu}^\nu$ and $A_\mu^{\ \ \nu}$ (defined by which index you raise), but if the tensor is symmetric $A_{\mu\nu}=A_{\nu\mu}$ you don't need to worry about position. To raise one of the indices on $g_{\mu\nu}$ requires the inverse metric. Is it clear if it write $g^\mu_\nu = g^{\mu\rho} g_{\rho\nu}$ and then set $\mu=\nu$ and sum?
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Michael BrownMar 14 '13 at 11:39